I've been scouring the web for explanations of Stimulated emission, and have seen questions Like this one on directions and frequencies, this one on energy conservation, this one one lasers and stimulated emission, and this question on explaining what stimulated emission is. I have found a lot of these unsatisfactory and I can't quite pinpoint why.

My current understand is as follows: I can start by imagining a two level system, $|e\rangle, | g \rangle$ with some difference in energy given by $\hbar \omega$. If my system starts in the excited state, an incident photon in resonance, thus with some energy $\hbar \omega$, can cause the system to drop to the ground state and release a photon with the same energy and in the same direction. I'm also still looking for a reference that quantizes the electric field and shows this, but in a semiclassical picture, a treatment of Time Dependent Perturbation theory seems to show that this can happen.

My question is: What is the intuitive reason for this process to occur, and how does the second photon know what direction to travel in?

I'm finding the explanation that "a perturbation can cause emission" to be unsatisfactory because it isn't obvious to me physically what part of the harmonic perturbation would give the photon a direction. I've seen explanations that tried to relate this to constructive interference in the field as well. There have been explanations that say a photon in resonance "shakes" the two-level system in resonance leading to emission, but I don't see the connection to the directionality of the emitted light. I guess I just don't understand why amplification should happen when a photon interacts with a two-level system in the excited state.

This entire line of questioning started when a friend asked me (at the atomic level) how a mirror works, and I've been finding more gaps in my knowledge on basic light-matter interactions!

  • $\begingroup$ You are barking up the wrong tree. The reason why it is coherent and going in the same direction is due to boson statistics, not where you are looking. Also, you need to specify what you mean by intuitive in this context. And a mirror works for a totally different reason from lasers. This has been a wild ride. $\endgroup$ Mar 3 at 22:12
  • 1
    $\begingroup$ There is no reason to expect your intuition (which is developed based on your experience with macro-scale objects) to give you correct expectations about quantum-scale objects. If there is an "intuitive" explanation for any quantum phenomenon, it's just by chance, not because our intuition can be used reliably to predict quantum behavior. $\endgroup$
    – The Photon
    Mar 3 at 23:09

1 Answer 1


There is an excellent classical analogy of this process, called “Three Wave Mixing”; it happens in plasmas and in non-linear optics.

In this situation, three different waves of three different frequencies interact in a nonlinear (but classical) manner to exchange energy between the modes. Suppose you have three waves with three frequencies, obeying the relation $\omega_e = \omega_g + \omega_p$. (This ‘frequency matching’ is usually a requirement for getting a strong three-wave interaction.) When these three waves interact, it turns out (because of something called the Manley-Rowe relations) that the energy $E_e$ gained/lost by the $\omega_e$ wave always is split up between the other two energies $E_g$ and $E_p$ in proportion to their frequencies. Specifically, $\Delta E_e/\omega_e = -\Delta E_g/\omega_g = -\Delta E_p/\omega_p$. So the number of "quanta" exchanged between these three modes just what you’d expect if they were made up of particles with energy $\hbar \omega$.

What’s more, if you think of $E_p$ like the energy in an electromagnetic mode, increasing the intensity of this mode will linearly increase the probability of exchanging energy. And the direction of energy exchange depends on where the energy starts. Start with the energy in $E_g$, and $E_p$ will drive it to $E_e$ (analgous to absorption). But start with the energy in $E_e$, and now you’ll get the stimulated emission analog; the presence of $E_p$ causes the energy to leave $E_e$ into both $E_p$ and $E_g$. (If all three are populated, the energy transfer direction depends on the relative phases of the oscillators.)

To finish the analogy, you need to think of the photons as single wave, not a set of discrete particles. That step is easy enough (classical EM waves are the classical analog of photons), but you also need to think of the excited and the ground states as two more waves, which may be a bit harder. (Maybe for this you need to think of a group of atoms as a box holding two standing electron waves, one for the excited state, and one for the ground state – or deBroglie oscillations around a Bohr atom.). Even if you don't like this last step, even if you just take the classical EM wave step, it becomes obvious why the “emitted photon” looks just like the incoming photon. They’re not really distinct at all, just different pieces of the total electromagnetic energy $E_p$ in a single EM wave. If $E_p$ increases for any reason, it will look as if new photons have appeared which have the same properties as the original wave.

(To see this analogy pushed even further, you might check out https://arxiv.org/abs/2208.05792 .)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.